Q.A stone of mass $m$ is tied to an elastic string of negligble mass and spring constant $k$. The unstretched length of the string is $L$ and has negligible mass. The other end of the string is fixed to a nail at a point $P$. Initially the stone is at the same level as the point P. The stone is dropped vertically from point P.

The key idea is that the motion has two distinct phases: free fall until the string becomes taut, followed by a spring-mass system where gravity and the spring force combine.

(a) The stone falls freely through distance $L$ (string slack). It then stretches the string by an additional distance $x$, coming to rest when the net work done by gravity and the spring force equals the change in kinetic energy (zero at start and at the lowest point).

From energy conservation:

$mg(L+x) = \frac12 k x^2$

Solving: $\frac12 k x^2 - mg x - mgL = 0$

$x = \frac{mg + \sqrt{m^2 g^2 + 2kmgL}}{k}$ (positive root).

Thus $y = L + x$.

(b) Maximum velocity occurs when the net force is zero during the spring phase: $mg = kx_0$, so $x_0 = mg/k$.

Energy from start to that point:

$\frac12 m v_{\max}^2 = mg(L + x_0) - \frac12 k x_0^2$

Substitute $x_0$: …

Concept and Intuition

This is a two‑phase problem. Until the stone has fallen a distance $L$, the string is slack — the stone is in free fall. Once the string begins to stretch, the stone experiences a linear restoring force $F = -k\,x$ (where $x$ is the extension beyond $L$), so it enters simple harmonic motion about the equilibrium stretch $x_0 = mg/k$.

The key insight: the stone’s lowest point is not the equilibrium point — it overshoots because of the kinetic energy gained during free fall. In SHM, the amplitude is determined by the total mechanical energy at the moment the string becomes taut.

Step‑by‑Step Solution

1. Free‑fall phase (string slack)

The stone falls from rest through a vertical distance $L$. Using $v^2 = u^2 + 2as$:

This is the speed just as the string becomes taut.

2. Define coordinates for the stretch phase

Let $y$ be the distance fallen from the nail. For $y > L$, the extension of the string is $x = y - L$. The net force on the stone (taking downward as positive) is:

The equilibrium position $y_e$ occurs when $F = 0$:

Let $x_0 = mg/k$ be the static stretch.

3. Lowest point (part a)

At the lowest point, the stone comes to rest instantaneously. Use energy conservation from the start (at the nail) to the lowest point $y_{\max}$.

• Initial energy (at $y=0$, $v=0$): $E_i = 0$ (taking gravitational PE = 0 at the nail)
• At lowest point $y = y_{\max}$: Gravitational PE: $-mg\,y_{\max}$ (since $y$ is downward, height lost is $y_{\max}$) Elastic PE: $\frac12 k (y_{\max} - L)^2$ Kinetic energy: $0$

Energy conservation:

Let $u = y_{\max} - L$ (the maximum extension). Then $y_{\max} = L + u$, and:

Multiply by 2:

Solve the quadratic:

Only the positive root is physical (extension is positive):

Thus the distance from the nail when the stone first comes to rest is:

4. Maximum velocity (part b)

Maximum velocity occurs at the equilibrium point $y_e = L + mg/k$, because that’s where the net force is zero and the acceleration changes sign.

Use energy conservation from the start to $y_e$:

At $y_e$:

Gravitational PE: $-mg\,y_e = -mg\left(L + \frac{mg}{k}\right)$

Elastic PE: $\frac12 k (y_e - L)^2 = \frac12 k \left(\frac{mg}{k}\right)^2 = \frac{m^2g^2}{2k}$

Kinetic energy: $\frac12 m v_{\max}^2$

Energy conservation ($E_i = 0$):

Simplify:

Multiply by $2/m$:

Thus:

$v_{\max} = \sqrt{2gL + \frac{mg^2}{k}}$ …

Phase 1: free fall for distance L, v0=sqrt(2gL). Phase 2: taut string, SHM about equilibrium y_e=L+mg/k. (a) lowest point from energy conservation: y_max=L+mg/k+sqrt((mg/k)^2+2mgL/k).

(b) max velocity at y_e: v_max=sqrt(2gL+mg^2/k). …